Nope, sure don't sesrch it on amazon or something. Maybe youll find one and be able to but it.
Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
Answer: Option (d) is the correct answer.
Explanation:
Electronegativity value of hydrogen is 2.2.
Electronegativity value of chlorine is 3.16.
Electronegativity value of carbon is 2.55.
Electronegativity value of oxygen is 3.44.
Electronegativity value of nitrogen is 3.04.
Electronegativity value of sodium is 0.93.
Electronegativity value of iodine is 2.66.
Therefore, calculate the electronegativity difference between the bonded atoms as follows.
- Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen
= 3.16 - 2.2
= 0.96
- Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon
= 3.44 - 2.55
= 0.89
- Electronegativity difference of
= Electronegativity value of nitrogen - electronegativity value of nitrogen
= 3.04 - 3.04
= 0
- Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium
= 2.66 - 0.93
= 1.73
So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.
Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.
Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,

The expression for rate of reaction for the reactant :
![\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNH_3%3D-%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DO_2%3D-%5Cfrac%7B1%7D%7B7%7D%5Ctimes%20%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The expression for rate of reaction for the product :
![\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DNO_2%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
From this we conclude that, all the options are correct.